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Viktor [21]
3 years ago
15

Q 16.1: Select the first step in the Wittig reaction. A : The alkyltriphenylphosphonium halide is deprotonated by a base to make

a phosphorus ylide. B : The phosphorus ylide reacts with the aldehyde or ketone to make an oxaphosphetane. C : Triphenylphosphine attacks the alkyl halide to produce an alkyltriphenylphosphonium halide. D : The oxaphosphetane decomposes to form the alkene and triphenylphosphine oxide.
Chemistry
1 answer:
Dennis_Churaev [7]3 years ago
6 0

Answer:

The phosphorus ylide reacts with the aldehyde or ketone to make an oxaphosphetane.

Explanation:

The Wittig reaction is a reaction that occurs between a phosphorus ylide and an aldehyde or ketone. The final products are an alkene and triphenyl phosphine oxide.

The first step in the reaction is the attack of the phosphorus ylide on the aldehyde or ketone. This is followed by attack of oxygen on phosphorus to form a [2+2] cycloaddition product (oxaphosphetane)  which  decomposes to form the alkene and triphenylphosphine oxide.

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What volume of DI water, in mL, must you use to dissolve 30.0 g of NaOH in order to make a 1.25 M solution? mL (round to whole n
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Answer:

600mL

Explanation:

Molarity of a solution (M) = number of moles (n) ÷ volume (V)

number of moles = mass/molar mass

Molar mass of NaOH = 23 + 16 + 1

= 40g/mol

mole = 30/40

n of NaOH = 0.75mol

Using Molarity = n/V

V = number of moles ÷ molarity

V = 0.75 ÷ 1.25

V = 0.6L

In milliliters (mL), the volume of NaOH will be 0.6 × 1000

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4 0
3 years ago
A laboratory analysis of a 100 g sample finds it is composed of 1.8 g hydrogen, 56.1 g sulfur, and 42.1 g oxygen. What is its em
Neporo4naja [7]

Answer: The empirical formula is H_2S_2O_3

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mas of H = 1.8 g

Mass of S = 56.1 g

Mass of O = 42.1 g

Step 1 : convert given masses into moles.

Moles of H =\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.8g}{1g/mole}=1.8moles

Mass of S =\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{56.1g}{32g/mole}=1.8moles

Moles of O=\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{42.1g}{16g/mole}=2.6moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For H = \frac{1.8}{1.8}=1

For S = \frac{1.8}{1.8}=1

For O =\frac{2.6}{1.8}=1.5

Converting to whole number ratios

The ratio of H: S: O= 2: 2: 3

Hence the empirical formula is H_2S_2O_3

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3 years ago
Which of the following substance is most likely basic​
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5 0
3 years ago
Menthol, the substance we can smell in mentholated cough drops, is composed of c, h, and o. a 9.045×10−2 −mg sample of menthol i
Ket [755]

Answer:

            Empirical Formula  =  C₁₀H₂₀O

Solution:

Data Given:

                      Mass of Menthol  =  9.045 × 10⁻² mg  =  9.045 × 10⁻⁵ g

                      Mass of CO₂  =  0.2546 mg  =  0.0002546 g

                      Mass of H₂O  =  0.1043 mg  =  0.0001043 g

Step 1: Calculate %age of Elements as;

                      %C  =  (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100

                      %C  =  (0.0002546 ÷ 9.045 × 10⁻⁵) × (12 ÷ 44) × 100

                      %C  =  (2.814) × (12 ÷ 44) × 100

                      %C  =  2.814 × 0.2727 × 100

                      %C  =  76.73 %


                      %H  =  (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100

                      %H  =  (0.0001043 ÷ 9.045 × 10⁻⁵) × (2.02 ÷ 18.02) × 100

                      %H  =  (1.153) × (2.02 ÷ 18.02) × 100

                      %H  =  1.153 × 0.1120 × 100

                     %H  =  12.91 %


                      %O  =  100% - (%C + %H)

                      %O  =  100% - (76.73% + 12.91%)

                      %O  =  100% - 89.64%

                     %O  =  10.36 %

Step 2: Calculate Moles of each Element;

                      Moles of C  =  %C ÷ At.Mass of C

                      Moles of C  = 76.73 ÷ 12.01

                     Moles of C  =  6.3888 mol


                      Moles of H  =  %H ÷ At.Mass of H

                      Moles of H  = 12.91 ÷ 1.01

                      Moles of H  =  12.7821 mol


                      Moles of O  =  %O ÷ At.Mass of O

                      Moles of O  = 10.36 ÷ 16.0

                      Moles of O  =  0.6475 mol

Step 3: Find out mole ratio and simplify it;

                C                                        H                                     O

            6.3888                              12.7821                            0.6475

     6.3888/0.6475                  12.7821/0.6475                 0.6475/0.6475

               9.86                                   19.74                                   1

             ≈ 10                                      ≈ 20                                     1

Result:

         Empirical Formula  =  C₁₀H₂₀O₁

8 0
3 years ago
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